【四边形或斜率优化】HDU 3480 Division

通道

题意：给出n个数字，要你把这n个数字分成m堆，每一堆的价值是(max(val) - min(val)) ^ 2 要你求出分成m堆之后得到的最小价值

思路：排序，dp[i][j]表示前j个数字，分成i堆的最小价值，dp[i][j] = min(dp[i - 1][k] + (val[j] - val[k + 1]) ^ 2) 。

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAX_N = 10007;

typedef int ll;

template <class T>
inline bool rd(T &ret) {
    char c; int sgn;
    if(c = getchar() , c == EOF) return false;
    while(c != '-' && (c < '0' || c > '9')) c = getchar();
    sgn = (c == '-') ? -1 : 1;
    ret = (c == '-') ? 0 : (c - '0');
    while(c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
    ret *= sgn;
    return true;
}

int n, m;
int s[MAX_N >> 1][MAX_N], a[MAX_N];
ll dp[MAX_N >> 1][MAX_N];

int main() {
    int T; scanf("%d", &T); int cas = 0;
    while (T-- > 0) {
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= n; ++i) rd(a[i]);
        sort(a + 1, a + 1 + n);
        memset(s, 0, sizeof s);
        for (int i = 1; i <= m; ++i) 
            for (int j = i; j <= n; ++j) {
                if (j <= i) dp[i][j] = 0;
                else dp[i][j] = 0x7f7f7f7f;
            }
        for (int i = 1; i <= n; ++i) s[1][i] = 1, dp[1][i] = (a[i] - a[1]) * (a[i] - a[1]);
        for (int i = 2; i <= m; ++i) s[i][n + 1] = n;
        for (int i = 2; i <= m; ++i) 
            for (int j = n; j >= i; --j) {
                for (int k = s[i - 1][j]; k <= s[i][j + 1]; ++k) {
                    int w = (a[j] - a[k + 1]) * (a[j] - a[k + 1]);
                    if (dp[i][j] > dp[i - 1][k] + w) {
                        dp[i][j] = dp[i - 1][k] + w;
                        s[i][j] = k;
                    }
                }
            }
        printf("Case %d: %d\n", ++cas, dp[m][n]);
    }
    return 0;
}

 

斜率优化：

对于第i个数，它要构成第j个集合时，假设它之前某两个数k1,k2,那么取k2为分割点比取k1为分割点优的条件就是：

    dp[k2][j - 1] + (a[i] - a[k2 + 1])^2 < dp[k1][j - 1] + (a[i] - a[k1 + 1])^2

=>  dp[k2][j - 1] + a[k2 + 1]^2 + a[i]^2 - 2 * a[i] * a[k2 + 1] < dp[k1][j - 1] + a[k1 + 1]^2 + a[i]^2 - 2 * a[i] * a[k1 + 1]

=>  dp[k2][j - 1] - dp[k1][j - 1] + a[k2 + 1]^2 - a[k1 + 1]^2 < 2 * a[i] * (a[k2 + 1] - a[k1 + 1])

代码：

include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int Case,n,m,head,tail;
int a[10050],dp[5050][10050],q[10010];
int work(){
   for (int i = 1;i <= n;i ++) dp[1][i] = (a[i] - a[1]) * (a[i] - a[1]);
   for (int i = 2;i <= m;i ++){
       head = tail = 0;
       q[tail ++] = i - 1;
       for (int j = i;j <= n;j ++){
           while (head + 1 < tail){
               int p1 = q[head],p2 = q[head + 1];
               int x1 = a[p1 + 1],x2 = a[p2 + 1];
               int y1 = dp[i - 1][p1] + x1 * x1,y2 = dp[i - 1][p2] + x2 * x2;
               if (y2 - y1 <= 2 * a[j] * (x2 - x1)) head ++;
               else break;
           }
           int k = q[head];
           dp[i][j] = dp[i - 1][k] + (a[j] - a[k + 1]) * (a[j] - a[k + 1]);
           while (head + 1 < tail){
               int p1 = q[tail - 2],p2 = q[tail - 1],p3 = j;
               int x1 = a[p1 + 1],x2 = a[p2 + 1],x3 = a[p3 + 1];
               int y1 = dp[i - 1][p1] + x1 * x1,y2 = dp[i - 1][p2] + x2 * x2,y3 = dp[i - 1][p3] + x3 * x3;
               if ((y3 - y1) * (x2 - x1) <= (y2 - y1) * (x3 - x1)) tail --;
               else break;
           }
           q[tail ++] = j;
       }
   }
   return dp[m][n];
}
int main(){
   int T;
   scanf("%d",&T);
   while (T --){
       scanf("%d%d",&n,&m);
       for (int i = 1;i <= n;i ++) scanf("%d",&a[i]);
       sort(a + 1,a + 1 + n);
       printf("Case %d: %d\n",++ Case,work());
   }
   return 0;
}